prove that if a series of functions converges uniformly in D...

Question

I met this question searching online for exercises but I can't seem to solve it.

Prove that if the series of functions $\sum_{n=1}^\infty f_n(x)$ converges uniformly in group D, then the sequence of functions $\{f_n(x)\}$ converges uniformly to $0$ in group D.

I tried coming up with some sort of connection but I am just stuck as nothing seems to connect.

Answer 1

We'll denote the partial sum as $S_n(x) = \sum_{k=1}^n f_n(x)$.

The infinite sum converges uniformly in D, so for every $\epsilon >0$ there exists a $N$ such that for every $n>N$: $|S_{n+1}(x)-S_{n}(x)| < \epsilon $ , for all $x\in D$. (when a converging series is $\frac\epsilon2$-close to it's limit, the difference between two consecutive terms can't be bigger than $\epsilon$)

Using the equality $S_{n+1}(x)-S_{n}(x) = f_{n+1}(x)$, we find that $f_n$ converges to $0$ for all $x\in D$. Because we used the same $N$ for all $x\in D$ we get that it converges uniformly.

Answer 2

We are assuming $S_n$ converges uniformly to some function $S,$ where $S_n = \sum_{k=1}^{n}f_k.$ It follows that both $S_n,S_{n+1} \to S$ uniformly. Therefore $S_{n+1}-S_n\to S-S = 0$ uniformly. Since $S_{n+1}-S_n = f_{n+1},$ we have the result.